Why is the input altered when the function runs on a different declared variable?

Question

In the code below, the operatorSequence function should run using the variable "input" and not "inputArray", however the inputArray ("a") is altered when the function is run.

What is causing this issue, and how can it be fixed so that the inputArray isn't changed?

//factorial in maths
function factorial(input) {
  if (input == 0) return 1;
  if (input > 0) return input * factorial(input - 1);
}
//slices an index from an array
function sliceIndex(array, index) {
  let temp = array.slice(0, index);
  temp.push.apply(temp, array.slice(index + 1));
  return temp;
}

//operator sequence
function operatorSequence(inputArray, operatorArray, det) {
  //the function should run on the "input" variable
  let input = inputArray;
  //det determines bracket position
  //according to the bracket position, two input indices are merged into one, and an operator is deleted
  while(operatorArray.length > 0) {
    let temp1 = factorial(operatorArray.length - 1);
    let temp2 = Math.floor(det / temp1);
    input[temp2] = operatorArray[temp2](input[temp2], input[temp2 + 1]);
    operatorArray = sliceIndex(operatorArray, temp2);
    input = sliceIndex(input, temp2 + 1);
    det -= temp1 * temp2;
  }
  return input[0];
}

const a = [1, 8, 3, 4];
let b = [(a, b) => a + b, (a, b) => a - b, (a, b) => a * b];
console.log(operatorSequence(a, b, 0));
console.log(a);
//24
//[9, 8, 3, 4]

`let input = inputArray;` does not create a copy. `let input = inputArray.slice();` does. — Sebastian Simon, Mar 31 '18 at 21:28
Also, avoid using variables with the same name as the function containing them, like in `let b = [(a, b) => a + b, (a, b) => a - b, (a, b) => a * b];`. — Sebastian Simon, Mar 31 '18 at 21:29
@Xufox, the last is an array of functions. the names do not matter to the outside used variable names. — Nina Scholz, Mar 31 '18 at 21:30

Nina Scholz · Answer 1 · 2018-03-31T21:38:21.100

let input = inputArray;

The assignment of an array to another variable does not produce a new array with values. Instead it uses the same object reference to the original array.

Later, you assign values

input[temp2] = operatorArray[temp2](input[temp2], input[temp2 + 1]);

to the array input as inputArray as the original array a.

You could work with a shallow copy of the array by using Array#slice:

let input = inputArray.slice();

Why is the input altered when the function runs on a different declared variable?

1 Answers1